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OG10 #226 S, U,V

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OG10 #226 S, U,V

by oquiella » Fri Oct 16, 2015 8:14 am
If s, u, and v are positive integers and 2^s=2^u+ 2^v, which of the following must be true?


I. s=u
II. u does not = v
III. s>v


A. None
B. I only
C. II only
D. III only
E. II and III

quick approach?
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by [email protected] » Fri Oct 16, 2015 9:52 am
Hi oquiella,

Since you ask for a 'quick approach', I'm going to give you some hints so that you can try this question again on your own:

You would likely find it helpful to write down the first few 'powers of 2', so that you can physically see the numbers involved...

Note that the prompt 'restricts' us to POSITIVE INTEGERS for S, U and V.

2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 16

Now, using these numbers, and a little logic, which of following Roman Numerals can you prove are true (or not true)?

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by MartyMurray » Fri Oct 16, 2015 9:44 pm
oquiella wrote:If s, u, and v are positive integers and 2^s=2^u+ 2^v, which of the following must be true?


I. s=u
II. u does not = v
III. s>v


A. None
B. I only
C. II only
D. III only
E. II and III

quick approach?
Ok, I suspect you have figured this out by now. So in case you or anyone else is still interested here's a way bang this one out.

Notice that the question asks which "must be true."

So if we can come up with any example that does not fit a statement, then we can eliminate that statement.

I. s=u

This one is easy to eliminate. Clearly if v is a positive integer then s > u. So I is out.

II. u does not = v

We can do 2^2 = 2^1 + 2^1. So u can = v. So II is out.

III. s>v

Let's make u as low as possible. Given that s, u, and v are positive, the lowest possible value for u is 1.

So we have 2^s = 2 + 2^v. All the numbers are positive here. So without even doing any math really, we can tell that 2^s has to be greater than 2^v. So s has to be greater than v.

III is the only one that must be true and the correct answer is D.
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